Question

This is a problem from the CodingBat website. I am pasting the problem first and discussing my efforts after that:

Given two strings, base and remove, return a version of the base string where all instances of the remove string have been removed (not case sensitive). You may assume that the remove string is length 1 or more. Remove only non-overlapping instances, so with "xxx" removing "xx" leaves "x".

withoutString("Hello there", "llo") → "He there"
withoutString("Hello there", "e") → "Hllo thr"
withoutString("Hello there", "x") → "Hello there"

This is what I wrote so far:

public String withoutString(String base, String remove) {

   int len_b=base.length();
   int len_r = remove.length();
   String result="";

   if(len_b<1 || len_r<1)
   return "";

   for (int i =0;i<=len_b-len_r;i++)
   {
      if(base.substring(i,i+len_r).equals(remove))
      {
        i=i+len_r-1;
      }

      else
      { 
        result=result+base.substring(i,i+1);
      }  
   } 

   if(!(base.substring(len_b-len_r+1, len_b).equals(remove)))
   result=result+base.substring(len_b-len_r+1, len_b);

return result;
}

This passes all the test cases except for the ones where the removal of the string should be case-insensitive.

For example: withoutString("This is a FISH", "IS") → "Th a FH"

My code gives me "This is a FH" as I haven’t handled case sensitivity in my code. I know that with Regex this could be done in one line. I am more interested in knowing if there is a way to handle these kinds of test cases in my present code.
Also, please let me know if my code could be made more efficient/elegant.

Answer 1

String has an equalsIgnoreCase(String s) method.